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bài 1 mifk viết sai nha.
bài 1: cho A=1+3+3\(^2\)+3\(^3\)+...+3\(^{10}\).Tìm số tự nhiên n biết 2 x A + 1 = 3\(^n\)
B1:
\(A=1+3+3^2+3^3+...+3^{10}\\ 3A=3+3^2+3^3+3^4+...+3^{11}\\ 3A-A=3^{11}-1\\ \Rightarrow A=\frac{3^{11}-1}{2}\)
mấy câu khác tương tự nha
1: =>7/3x=3+1/3-8-2/3=-5-1/3=-16/3
=>x=-16/3:7/3=-7/16
2: =>1/3|x-2|=4/5+3/7=28/35+15/35=43/35
=>|x-2|=129/35
=>x-2=129/35 hoặc x-2=-129/35
=>x=199/35 hoặc x=-59/35
A=1−3+5−7+...+2001−2003+2005S=1−3+5−7+...+2001−2003+2005
=(1−3)+(5−7)+...+(2001−2003)+2005=(1−3)+(5−7)+...+(2001−2003)+2005(Có 1002 cặp)
=(−2).1002+2005=(−2).1002+2005
=−2004+2005=−2004+2005
=1
a)(25.5-52.2):(5.2)-3
= (25.5-25.2):10-3
= 25.(5-2):10-3
= 25.3:10-3
=75:10-3=7,5-3=4,5
b)(6.52 -137).2-23.(7+3)(Sai đề)
c)23-53 :52 +12.22
= 8-125:25+12.4
= 8-5+12.4=8-5+48=3+48=51
d)2.[(95+52:5):22 +180] -22.102
= 2.[(95+25:5):4+180]-4.100
= 2.[(95+5):4+180]-400
= 2.(100:4+180)-400
= 2. (25+180)-400
= 2. 205-400
= 410-400=10
e)27.22+54:53.24-3.25
= 128+625:125.24-3.32
= 128+5.24-96
= 128+120-96
= 248-96=152
f)2.[(7-33 :32):22+99]-100
=2.[(7-27:9):4+99]-100
=2.[(7-3):4+99]-100
=2. (4:4+99)-100
=2. (1+99)-100
=2. 100-100
= 200-100
=100
Chúc Bạn Học Tốt ^_^
a )
\(\frac{-4}{9}.\frac{1}{3}-\frac{4}{9}.\frac{5}{6}+\frac{3}{7}.\frac{4}{9}\)
\(=\frac{4}{9}.\left(-\frac{1}{3}-\frac{5}{6}+\frac{3}{7}\right)\)
\(=\frac{4}{9}.\left(-\frac{14}{42}-\frac{35}{42}+\frac{18}{42}\right)\)
\(=\frac{4}{9}.\frac{-31}{42}\)
\(=-\frac{62}{189}\)
b )
\(\frac{2}{3}:\frac{3}{7}-\frac{2}{3}:\frac{4}{3}+\frac{2}{3}:\frac{1}{21}\)
\(=\frac{2}{3}.\frac{7}{3}-\frac{2}{3}.\frac{3}{4}+\frac{2}{3}.21\)
\(=\frac{14}{9}-\frac{1}{2}+14\)
\(=\frac{28}{18}-\frac{9}{18}+14\)
\(=\frac{19}{18}+14\)
\(=1+14+\frac{1}{18}\)
\(=15\frac{1}{18}\)
c )
\(\left(5\frac{1}{3}+3\frac{2}{3}\right)-4\frac{1}{3}\)
\(=\left(5+3-4\right)+\left(\frac{1}{3}+\frac{2}{3}-\frac{1}{3}\right)\)
\(=4\frac{2}{3}\)
\(=\frac{14}{3}\)
a) \(-\frac{4}{9}\cdot\frac{1}{3}-\frac{4}{9}\cdot\frac{5}{6}+\frac{3}{7}\cdot\frac{4}{9}\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{1}{3}+\left(-\frac{4}{9}\right)\cdot\frac{5}{6}-\left(-\frac{4}{9}\right)\cdot\frac{3}{7}\)
\(=\left(-\frac{4}{9}\right)\left(\frac{1}{3}+\frac{5}{6}-\frac{3}{7}\right)\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{31}{42}=-\frac{62}{189}\)